First-Law Bookkeeping Around a Rectangular T–V Cycle: 21-Mat-A1 Thermodynamics, December 2017, Q1

Question 1: First-Law Bookkeeping Around a Rectangular T–V Cycle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $n=1$ mol monatomic ideal gas ($C_v=\tfrac32R=12.47\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $C_p=\tfrac52R=20.79\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$, $R=8.314\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}$). Step 1 (isothermal, $T=300$ K): A(10 L)→B(50 L). Step 2 (isochoric, $V=50$ L): B(300 K)→C(600 K). Step 3 (isothermal, $T=600$ K): C(50 L)→D(10 L). Step 4 (isochoric, $V=10$ L): D(600 K)→A(300 K).

Find. $P_A,P_B,P_C,P_D$; and $q$, $w$ (on the system), $\Delta E\,(=\Delta U)$ and $\Delta H$ for each of the four steps.

Figure 1 — The rectangular cycle in the T–V plane: Step 1 (A→B) and Step 3 (C→D) are isothermal legs; Step 2 (B→C) and Step 4 (D→A) are isochoric legs.

Approach. Get every corner pressure from $PV=nRT$ (noting 1 kPa·L $=$ 1 J, so $R=8.314$ kPa·L K⁻¹mol⁻¹), then apply $w=-nRT\ln(V_f/V_i)$ with $\Delta E=\Delta H=0$ on the isothermal legs, and $w=0$ with $q=nC_v\Delta T$, $\Delta H=nC_p\Delta T$ on the isochoric legs.

(a) Corner pressures. $P=\dfrac{nRT}{V}$, with $R=8.314$ kPa·L K⁻¹mol⁻¹ giving $P$ directly in kPa: $$P_A=\dfrac{(8.314)(300)}{10}=\boxed{249.4\ \text{kPa}},\qquad P_B=\dfrac{(8.314)(300)}{50}=\boxed{49.9\ \text{kPa}},$$ $$P_C=\dfrac{(8.314)(600)}{50}=\boxed{99.8\ \text{kPa}},\qquad P_D=\dfrac{(8.314)(600)}{10}=\boxed{498.8\ \text{kPa}}.$$
(b) Step 1 (A→B, isothermal, $T=300$ K). For an ideal gas $U$ and $H$ depend on $T$ alone, so $\Delta E=\Delta H=0$. The reversible isothermal work done ON the system is $$w=-nRT\ln\!\left(\dfrac{V_B}{V_A}\right)=-(8.314)(300)\ln\!\left(\dfrac{50}{10}\right)=-(2494.2)(1.6094)=\boxed{-4014.3\ \text{J}}.$$ With $\Delta E=0=q+w$, $q=-w=\boxed{+4014.3\ \text{J}}$ (heat absorbed to keep $T$ constant as the gas expands and does work on the surroundings).
(c) Step 2 (B→C, isochoric, $V=50$ L). $\Delta T=T_C-T_B=300$ K and $w=0$ (no volume change): $$\Delta E=q=nC_v\Delta T=(12.47)(300)=\boxed{+3741.3\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(300)=\boxed{+6235.5\ \text{J}}.$$
(d) Step 3 (C→D, isothermal, $T=600$ K). Again $\Delta E=\Delta H=0$, and the gas is now compressed ($V_D
(e) Step 4 (D→A, isochoric, $V=10$ L). $\Delta T=T_A-T_D=-300$ K and $w=0$: $$\Delta E=q=nC_v\Delta T=(12.47)(-300)=\boxed{-3741.3\ \text{J}},\qquad \Delta H=nC_p\Delta T=(20.79)(-300)=\boxed{-6235.5\ \text{J}}.$$ Check: summing all four legs gives $\Delta E_{cyc}=0+3741.3+0-3741.3=0$ and $\Delta H_{cyc}=0$, as required for a state function returning to its start. ✓

Quantity	Value
$P_A,P_B,P_C,P_D$	249.4, 49.9, 99.8, 498.8 kPa
Step 1 (A→B): q, w, ΔE, ΔH	+4014.3 J, −4014.3 J, 0, 0
Step 2 (B→C): q, w, ΔE, ΔH	+3741.3 J, 0, +3741.3 J, +6235.5 J
Step 3 (C→D): q, w, ΔE, ΔH	−8028.5 J, +8028.5 J, 0, 0
Step 4 (D→A): q, w, ΔE, ΔH	−3741.3 J, 0, −3741.3 J, −6235.5 J

Concept & Theory

Topic. First law of thermodynamics applied leg-by-leg to a rectangular ideal-gas T–V cycle.

Key relations. $PV=nRT$; $\Delta U=nC_v\Delta T$; $\Delta H=nC_p\Delta T$; $w_{on}=-nRT\ln(V_f/V_i)$ (isothermal); $w_{on}=0$ (isochoric); $\Delta U=q+w_{on}$.

Why this works. Because $U$ and $H$ of an ideal gas depend on temperature alone, the two isothermal legs (1 and 3) contribute nothing to $\Delta E$ or $\Delta H$ — all of the heat exchanged on those legs is converted directly to/from boundary work. The two isochoric legs (2 and 4) do no work at all, so their heat exchange shows up entirely as a change in internal energy. Over the closed cycle both state functions return to zero.

Common pitfall. Forgetting the sign convention on $w$: taking $w$ as work done BY the gas (positive on expansion) instead of work done ON the gas (negative on expansion) flips every sign in the table; this solution consistently reports $w$ as work done ON the system, matching $\Delta U=q+w$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 3–4 (first law, ideal-gas state functions and isothermal/isochoric work).

Question 1 of 7: First-Law Bookkeeping Around a Rectangular T–V Cycle

Question 1: First-Law Bookkeeping Around a Rectangular T–V Cycle (20 marks)